let $ \ \ 0

Question

let $ \ \ 0<a \leq b \leq c \in \mathbb{R}$ then prove that :

$$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$

I do not know where to start please help me !

How did you get 2 upvotes if you didn't show any effort? You don't have other accounts where you upvote a question of your own? — Anonymous196, Oct 28 '17 at 10:59
That's more a recreational type of question whereas this question is more like a contest inequality problem. The questioner should have posted some efforts even if it was leading nowhere. — Anonymous196, Oct 28 '17 at 11:41

Michael Rozenberg · Answer 1 · 2017-10-28T11:13:00.297

2

Your inequality is true!

We need to prove that $$\frac{a+b+c}{3}-\frac{3abc}{ab+ac+bc}\geq\frac{(c-a)^2}{6c}$$ or $$\frac{\sum\limits_{cyc}c(a-b)^2}{ab+ac+bc}\geq\frac{(c-a)^2}{2c}$$ or $$2c(a-b)^2+2a(b-c)^2\geq\left(\frac{ab}{c}+a-b\right)(c-a)^2.$$ Now, by C-S $$2c(a-b)^2+2a(b-c)^2=2ac\left(\frac{(b-a)^2}{a}+\frac{(c-b)^2}{c}\right)\geq$$ $$\geq\frac{2ac(b-a+c-b)^2}{a+c}=\frac{2ac(c-a)^2}{a+c}.$$ Thus, it remains to prove that $$2ac\geq(a+c)\left(\frac{ab}{c}+a-b\right)$$ or $$c^2a+c^2b\geq a^2b+a^2c,$$ which is obvious.

edited Oct 28 '17 at 11:13

answered Oct 28 '17 at 11:06

Michael Rozenberg

194,933

I did not understand that:$$2c(a-b)^2+2a(b-c)^2=2ac\left(\frac{(b-a)^2}{a}+\frac{(c-b)^2}{c}\right)\geq$$ $$\geq\frac{2ac(b-a+c-b)^2}{a+c}=\frac{2ac(c-a)^2}{a+c}$$ – Almot1960 Oct 28 '17 at 13:52
@Almot1960 It's Cauchy-Schwarz inequality. For positives $b_i$ we have: $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$ – Michael Rozenberg Oct 28 '17 at 15:53

score 1 · Answer 2 · answered Oct 28 '17 at 11:10

Clearing the denominators and moving all terms to the left we have to prove that $$-a^3b-a^3c+3a^2bc+4a^2c^2+2ab^2c-11abc^2+ac^3+2b^2c^2+bc^3\geq 0$$ Setting $$b=a+u,c=a+u+v$$ with $$u,v\geq 0$$ this inequality is equivalent to $$a^2(u-v)^2+4au^3+4au^2v+2auv^2+2av^3+3u^3+7u^3v+5u^2v^2+uv^3\geq 0$$ which is true.

let $ \ \ 0

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